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Question Number 144049 by physicstutes last updated on 20/Jun/21

$$\mathrm{Given}\mathrm{the}\mathrm{equation}1000=2000\left(\frac{1-{(1+t)}^{-n}}{t}\right)$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}t.$$

Answered by Olaf_Thorendsen last updated on 21/Jun/21

$$1000=2000\left(\frac{1-{(1+t)}^{-n}}{t}\right)$$$$\frac{1-{(1+t)}^{-n}}{t}=\frac{1}{2}$$$$\frac{1}{1+t}.\frac{1-{\left(\frac{1}{1+t}\right)}^n}{1-\frac{1}{1+t}}=\frac{1}{2}$$$$q(1-q^n)=\frac{1}{2}(1-q)$$$$\mathrm{Let}q=\frac{1}{1+t}$$$$q^{n+1}-\frac{3}{2}q+\frac{1}{2}=0$$

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